Given an oriented manifold $M$ and an oriented submanifold $\phi:N\to M$ we can obtain a homology class $\phi_*[N]\in H_*(M)$ where $[N]$ is the fundamental class of $N$. In general, it is not true that every homology class of $M$ can be represented by a submanifold in this manner, however for some special cases it is.

For example, for $M$ an oriented (and closed maybe?) 4-manifold every homology class can be represented by a submanifold. Another example is when $M$ an Euclidean configuration space.

My questions are:

1) Under what circumstances can every homology class of $M$ be represented by a submanifold and

2) What are some examples of manifolds who have homology classes not representable in this manner?

Yes, duplicate question and a question that has very good and well known answers in the literature.
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Ryan BudneyApr 13 '10 at 6:43

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Well it's not exactly the same question. The other question asks what is generated by images of fundamental classes of submanifolds. This question asks what is the image of the fundamental classes of submanifolds.
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Don StanleyApr 13 '10 at 6:59

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The original post seems to be closer to what I was looking for. Thanks for the link and sorry for the repost.
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SteveApr 13 '10 at 12:43

3 Answers
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Every class in $H_{n-1}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to S^1$ representing the Poincare dual in $H^1(M;Z)=[M,S^1]$ and take the preimage of a point. In dimensions>2 it can be taken connected.

Similarly, every class in $H_{n-2}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to CP^\infty$ representing the Poincare dual in $H^2(M;Z)=[M,CP^\infty]$, homotop $f$ into a finite skeleton, say $CP^N$, and take the preimage of $CP^{N-1}$.

Transversality says that if you can represent $x\in H_k(M)$ by a map of a smooth manifold (e.g. elements in the image of the Hurewicz map, or by Thom) , and $2k < n$, then you can represent it by an embedded submanifold (as Andy mentions above). For example, any class in $H_1(M)$ for $dim(M)\ge 3$. With care you can also make this work for $2k=n$, and there are techniques available in the "metastable" range (no triple points) involving generalizations of Whitney's trick and other ways to replace double points.

A weaker question replaces "homology class of an embedded submanifold" with $f_{\ast}([M])$ for some compact smooth manifold $M$ and an arbitrary continuous map $f:M \rightarrow X$. Once you give up looking at embedded submanifolds, there is also no reason to restrict yourself to $X$ being a manifold.

A lot was proven about this by Thom in his classic paper "Quelques propriétés globales des variétés différentiables", which is more famous for containing his work on cobordism theory. A few of the results contained in that paper are as follows.

1) Every mod $2$ homology class can be so represented.

2) Integrally, it is true for every class in $H_k$ for $k \leq 6$.

3) For every $k \geq 7$, there exist polyhedra $X$ and classes in $H_k(X)$ that cannot be so represented.

EDIT : One should also remark that the above is germane to the original question too in many cases. Namely, if $X$ is a smooth $n$-manifold and $M$ is a compact smooth $k$-manifold and $f:M \rightarrow X$ is arbitrary, then $f$ is homotopic to an embedding as long as $k < n/2$.

If you really want a submanifold then I guess you can't always do it. For a closed manifold $M$ consider two times the fundamental class $2[M]$. It's easy to see you can't represent this class as a submanifold when $M=S^1$. Perhaps if you take any class in $a\in H_*(M)$ with nonzero self intersection, then $2a$ can't be represented as a submanifold?

About your last sentence: as Paul notes, classes of codim 1 or 2 are always representable. In the codim one case, to double the homology class, compose a map to $S^1$ with the double covering of $S^1$ by itself.
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Tim PerutzApr 13 '10 at 13:46

@Don: It depends on what you mean by "self-intersection" and whether you allow disconnected submanifolds. eg if you take twice the first generator of H1(T), then 2x\cdot 2x=0 but the curve t\mapsto (e^{4\pi i t},1) has self-intersection 1. But 2 parallel copies of x is an embedded, disconnected manifold that represents 2x
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PaulApr 13 '10 at 14:17